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GREYUIT [131]
3 years ago
11

The attractive electrostatic force between the point charges 6.14×10−6 c and q has a magnitude of 0.845 n when the separation be

tween the charges is 9.26 m . part a find the sign and magnitude of the charge q.
Physics
1 answer:
Nataly [62]3 years ago
3 0
Coulomb's law states that F = k(q1)(q2)/r^2, where k = 9 x 10^9 N-m^2/C^2
Substituting the given values:

0.845 N = (9x10^9 N-m^2/C^2)(6.14x10^-6 C)(q) / (9.26^2)
q = 0.00131 = 1.31 x 10^-3 C

Since the force is attractive, the point charges must be of opposite signs, and q is negatively charged (-1.31 x 10^-3 C).
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Answer:

True

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3 years ago
__ is the second most abundant element in Earth’s crust. It is found in ___, and ___; and in _ which is used to make pottery.
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silicon is the second most abundant element in Earth’s crust. It is found in the sun and stars; and in clay which is used to make pottery.

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6 0
3 years ago
A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
Read 2 more answers
An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v
maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

T = \frac{PV}{nR}

T_{i} = T_{f}

\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}

Where:

P_{i}: is the initial pressure = 5.79 atm

P_{f}: is the final pressure =?

V_{i}: is the initial volume = 420 cm³

V_{f}: is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!        

3 0
3 years ago
A manganese atom is pictured below.
Rufina [12.5K]
Manganese   has  2 (two) electron  that   would  free  floating   and   able  to  form  a  metallic  bond.
  The    electronic  configuration  of  manganese  is  (Ar)  3d5 4s2.  The   two   electron  in  4s  orbital  are  the  valence    electron  which  can  freely  move  from  one  place  to  another.
5 0
3 years ago
Read 2 more answers
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